Re: [PATCH] bpf: task_group_seq_get_next: cleanup the usage of next_thread()

Oleg Nesterov
2023-08-21T18:35:41Z

On 08/21, Kui-Feng Lee wrote:
>
>
> On 8/21/23 08:09, Oleg Nesterov wrote:
> >1. find_pid_ns() + get_pid_task() under rcu_read_lock() guarantees that we
> > can safely iterate the task->thread_group list. Even if this task exits
> > right after get_pid_task() (or goto retry) and pid_alive() returns 0 >
> > Kill the unnecessary pid_alive() check.
>
> This function will return next_task holding a refcount, and release the
> refcount until the next time calling the same function. Meanwhile,
> the returned task A may be killed, and its next task B may be
> killed after A as well, before calling this function again.
> However, even task B is destroyed (free), A's next is still pointing to
> task B. When this function is called again for the same iterator,
> it doesn't promise that B is still there.

Not sure I understand...

OK, if we have a task pointer with incremented refcount and do not hold
rcu lock, then yes, you can't remove the pid_alive() check in this code:

rcu_read_lock();
if (pid_alive(task))
do_something(next_thread(task));
rcu_read_unlock();

because task and then task->next can exit and do call_rcu(delayed_put_task_struct)
before we take rcu_read_lock().

But if you do something like

rcu_read_lock();

task = find_task_in_some_rcu_protected_list();
do_something(next_thread(task));

rcu_read_unlock();

then next_thread(task) should be safe without pid_alive().

And iiuc task_group_seq_get_next() always does

rcu_read_lock(); // the caller does lock/unlock

task = get_pid_task(pid, PIDTYPE_PID);
if (!task)
return;

next_task = next_thread(task);

rcu_read_unlock();

Yes, both task and task->next can exit right after get_pid_task(), but since
can only happen after we took rcu_read_lock(), delayed_put_task_struct() can't
be called until we drop rcu lock.

What have I missed?

Oleg.